Question

The positive value of \[k\] for which the equations \[x^{2} + kx + 64 = 0\] and \[x^{2} – 8x + k = 0\] will both have real roots is ______
A. \[4\]
B. \[8\]
C. \[12\]
D. \[16\]

Answer 1

Hint:In this question, we need to find the value of k in both the equations \[x^{2} + kx + 64 = 0\] and \[x^{2} – 8x + k = 0\] . And also given that both the equations will have real roots. For finding the root of both the equations, we will use the concept of discriminant of the quadratic equations. The discriminant is nothing but the part of the quadratic formula underneath the square root symbol \[b^{2} – 4ac\]. Using this we can find the value of \[k\].

Complete step by step answer:
Given, \[x^{2} + kx + 64 = 0\] and \[x^{2} – 8x + k = 0\]. Here we need to find the value of \[k\]. Let us know the concept of discriminant of the quadratic equations. Mathematically, a discriminant is nothing but a function of the coefficients of a polynomial equation that expresses the nature of the roots of the given quadratic equation. We know that the standard form of the equation is \[ax^{2} + bx + c = 0\]. Then the discriminant of the equation is \[b^{2} – 4ac\], Where \[a\] is the coefficient of \[x^{2}\] , \[b\] is the coefficient of \[x\] and \[c\] is a constant term.The condition for a quadratic equation to have real roots is that the discriminant of the equations must be greater than or equal to zero.

First let us consider the first equation.
\[x^{2} + kx + 64 = 0\]
Here \[a\] is \[1\] , \[b\] is \[k\] and \[c\] is \[64\]
Discriminant , \[b^{2} – 4ac \geq 0\]
\[\Rightarrow \ k^{2} – 4\left( 1 \right)\left( 64 \right) \geq 0\]
On simplifying,
We get,
\[\Rightarrow \ k^{2} – 256 \geq 0\]
We can rewrite \[256\] as \[16 \times 16\] . Since \[256\] is a perfect square number.
\[\Rightarrow \ k^{2} – 16^{2} \geq 0\]
We know that
\[\left( a^{2} – b^{2} \right) = \left( a – b \right)\left( a + b \right)\]

By rewriting we get,
\[\Rightarrow \ (k + 16)(k – 16)\ \geq 0\]
Which can be written as \[(k + 16) \geq 0\] and \[(k – 16) \geq 0\]
Thus we get \[k \geq 16\] and \[k \leq – 16\]
Hence \[k \geq 16\] (since \[k\] is positive)
Now let us consider the second equation.
\[x^{2} – 8x + k = 0\]
Here \[a\] is \[1\] , \[b\] is \[- 8\] and \[c\] is \[k\].
Discriminant , \[b^{2} – 4ac \geq 0\]
\[\Rightarrow \ \left( - 8 \right)^{2} – 4\left( 1 \right)\left( k \right) \geq 0\]

On simplifying we get,
\[\Rightarrow \ 64 – 4k \geq 0\]
On subtracting \[- 64\] on both sides we get,
\[\Rightarrow \ - 4k \geq – 64\]
Now on dividing \[- 4\] on both sides,
We get,
\[\Rightarrow \ k \leq \dfrac{- 64}{- 4}\]
On simplifying we get,
\[\therefore \ k \leq 16\]
Therefore we get the value of \[k\] as \[16\] that satisfies both the conditions.

Hence, the correct answer is option D.

Note:To solve these types of questions, we should have a strong grip over the quadratic equations. We also need to know that when the discriminant value of the equation is positive, we get two real solutions for the equation. Also when the discriminant value of the equation is zero, we get one real solution and finally when the discriminant value of the equation is negative, we get a pair of complex solutions. The discriminant of the quadratic equations reveals the nature of the roots of the quadratic equation. It also helps us to determine the solution of an equation.